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Is xy > (x^2)(y^2)?

This topic has 3 expert replies and 0 member replies

Top Member

Is xy > (x^2)(y^2)?

Post Sun Jan 15, 2017 5:13 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Is xy > (x^2)(y^2)?
    (1) 4x^2 = 19
    (2) y^2 = 1

    OA:C

    Source:Math Revolution

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    Post Sun Jan 15, 2017 5:28 pm
    When DS questions involve algebra in the question stem, always try to simplify the algebra before looking at the statements.

    With xy > (x^2)(y^2), it's tempting to want to divide both sides by xy. But what if one of the variables is 0? Or if xy is negative, we'd have to flip the sign. Instead, think conceptually: When would a product of two numbers be greater than the product of the squares? xy would have to be non-negative, since the squares will always be non-negative. It would also have to be between 0 and 1, because only numbers in that range get smaller when they are squared.

    Target question: is xy between 0 and 1?

    (1) 4x^2 = 19
    This tells us that x^2 = 4.75, so x is the positive or negative square root of 4.75 (something close to 2 or -2). This doesn't tell us about y, so it's insufficient.

    (2) y^2 = 1
    This tells us that y = 1 or -1, but tells us nothing about x. Insufficient.

    1 & 2 together
    It seems like we don't have enough information, because we don't know whether x and y are positive or negative. But - it doesn't matter. The product xy will equal (approximately) 2 or -2. Either way, (xy)^2 will be larger than xy. Sufficient.

    The answer is C.

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    Post Sun Jan 15, 2017 5:34 pm

    _________________


    Ceilidh Erickson
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    EdM in Mind, Brain, and Education
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    Thanked by: gmatdestroyer13
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    Post Sun Jan 15, 2017 10:13 pm
    NandishSS wrote:
    Is xy > (x^2)(y^2)?
    (1) 4x^2 = 19
    (2) y^2 = 1

    OA:C

    Source:Math Revolution
    We have a situation: Is xy > (xy)^2?

    Whether one/both of x and y is/are positive/negative, RHS [(xy)^2] is positive. So we must focus our energy on whether xy is positive and is greater than (xy)^2 or not.

    S1:
    4x^2 = 19 => x =(19/4)^(1/2) => x can be positive/negative
    We do not have any information about y. S1 itself is insufficient.

    S2:
    y^2 = 1 => y = +/-1.
    We do not have any iformation about x. S2 itself is insufficient.

    S1 and S2:
    By plugging in the value of x and y, we get,

    [+/-(19/4)^(1/2)]*[+/-1] > [19/4]*[1]?

    If we take the product of x and y positive, then

    |(19/4)^(1/2)| > 19/4?, the answer is NO. The square root of a number greater than 1 (19/4 > 1) is less than the number.

    If we take the product of x and y negative, then

    -|(19/4)^(1/2)| > 19/4?, the answer is NO. LHS is negative and RHS is positive.

    Sufficient.

    Answer: C

    -Jay
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